If you set up the data in one long column with A and B as a new column, you then can run your regression model as a GLM with a continuous time variable and a nominal "experiment" variable (A, B). The output of the ANOVA will give you the significance of the difference between the parameters. "intercept' is the common intercept and the "experiment" factor will reflect differences between the intercepts (actually overall means) between the experiments. the "Time" factor will be the common slope, and the interaction is the difference between the experiments with respect to the slope.
I have to admit I cheat (?) and run the models separately first to get the two sets of parameters and their errors and then run the combined model to acquire the differences between the treatments (in your case A and B)...
For a general GLM, the deviance $\Delta$ is defined as $\Delta:=2(\tilde{l}-\hat{l})$ where $\tilde{l}$ and $\hat{l}$ are the loglikelihood of the saturated and our model, respectively.
The general use of the deviance in goodness-of-fit testing for a GLM, with $n$ observations and $p$ parameters in the model, uses that the deviance is approximately chi-squared distributed with $n-p$ degrees of freedom, i.e $\Delta \sim \chi^2_{n-p}$.
A large deviance would indicate that our model is "far" from the perfect fitting one. Thus a large value of the deviance would indicate a poorly fitting model. So $P(\Delta<\chi^2_{n-p})$ should be small.
However, for the binomial response distribution in a GLM, the deviance is not always a good measure of fit.
If your data is, or can be grouped, the chi-square approximation will work if both $n_i \hat{\pi_i}>5$ and $n_i(1-\hat{\pi_i})>5$ for each group $i$.
But, if you have binary responses, i.e $y_i$ either 0 or 1, the chi-square approximation will not be correct. Also the deviance will be connected to the actual responses only through the fitted values. How can you assess a goodness of fit with an expression for the deviance only containing estimated values?
A good alternative if you need a test is the Hosmer-Lemeshow test.
When it comes to measuring your models actual prediction power, you could try to make a ROC-curve. It will give you an indication of how well your model is performing.
We can define sensitivity as the relative frequency of predicting an event when an event takes place, i.e guessing right when $y_i=1$, and specificity as the relative frequency of predicting a non-event when there is no event, i.e guessing right when $y_i=0$. Ideally they are both close to 1.
If we estimate our model, calculate the probability for each observation, and according to some threshold value classify it as either an event, or a non-event, we can calculate the sensitivity and specificity of our model at this threshold. A threshold of zero, would yield sensitivity of 0 and specificity of 1, a threshold of 1 yields sensitivity of 1 and specificity of 0. So every ROC-curve starts in (0,0) and ends in (1,1) (as we have 1-specificity on the x-axis). Plotted below is such a curve.
A model that predicts well will have a sharply rising ROC-curve yielding a high sensitivity and high specificity, the further a curve is toward the top left, the better its predicting power. A model with a ROC-curve corresponding to the $45^{\circ}$-line is a model no better than simply guessing.
In conclusion, the deviance is not a good measure of fit in a binary regression GLM. If you really need a test, use the Hosmer-Lemeshow. If you're interested in the actual prediction capabilities of the model, use an ROC curve. There are several packages in R that will do this for you, pROC is one.
Best Answer
Your question may come from the fact that you are dealing with Odds Ratios and Probabilities which is confusing at first. Since the logistic model is a non linear transformation of $\beta^Tx$ computing the confidence intervals is not as straightforward.
Background
Recall that for the Logistic regression model
Probability of $(Y = 1)$: $p = \frac{e^{\alpha + \beta_1x_1 + \beta_2 x_2}}{1 + e^{ \alpha + \beta_1x_1 + \beta_2 x_2}}$
Odds of $(Y = 1)$: $ \left( \frac{p}{1-p}\right) = e^{\alpha + \beta_1x_1 + \beta_2 x_2}$
Log Odds of $(Y = 1)$: $ \log \left( \frac{p}{1-p}\right) = \alpha + \beta_1x_1 + \beta_2 x_2$
Consider the case where you have a one unit increase in variable $x_1$, i.e. $x_1 + 1$, then the new odds are
$$ \text{Odds}(Y = 1) = e^{\alpha + \beta_1(x_1 + 1) + \beta_2x_2} = e^{\alpha + \beta_1 x_1 + \beta_1 + \beta_2x_2 } $$
$$ \frac{\text{Odds}(x_1 + 1)}{\text{Odds}(x_1)} = \frac{e^{\alpha + \beta_1(x_1 + 1) + \beta_2x_2} }{e^{\alpha + \beta_1 x_1 + \beta_2x_2}} = e^{\beta_1} $$
Log Odds Ratio = $\beta_1$
Relative risk or (probability ratio) = $\frac{ \frac{e^{\alpha + \beta_1x_1 + \beta_1 + \beta_2 x_2}}{1 + e^{ \alpha + \beta_1x_1 + \beta_1 + \beta_2 x_2}}}{ \frac{e^{\alpha + \beta_1x_1 + \beta_2 x_2}}{1 + e^{ \alpha + \beta_1x_1 + \beta_2 x_2}}}$
Interpreting coefficients
How would you interpret the coefficient value $\beta_j$ ? Assuming that everything else remains fixed:
Confidence intervals for a single parameter $\beta_j$
Since the parameter $\beta_j$ is estimated using Maxiumum Likelihood Estimation, MLE theory tells us that it is asymptotically normal and hence we can use the large sample Wald confidence interval to get the usual
$$ \beta_j \pm z^* SE(\beta_j)$$
Which gives a confidence interval on the log-odds ratio. Using the invariance property of the MLE allows us to exponentiate to get $$ e^{\beta_j \pm z^* SE(\beta_j)}$$
which is a confidence interval on the odds ratio. Note that these intervals are for a single parameter only.
If you include several parameters you can use the Bonferroni procedure, otherwise for all parameters you can use the confidence interval for probability estimates
Bonferroni procedure for several parameters
If $g$ parameters are to be estimated with family confidence coefficient of approximately $1 - \alpha$, the joint Bonferroni confidence limits are
$$ \beta_g \pm z_{(1 - \frac{\alpha}{2g})}SE(\beta_g)$$
Confidence intervals for probability estimates
The logistic model outputs an estimation of the probability of observing a one and we aim to construct a frequentist interval around the true probability $p$ such that $Pr(p_{L} \leq p \leq p_{U}) = .95$
One approach called endpoint transformation does the following:
Since $Pr(x^T\beta) = F(x^T\beta)$ is a monotonic transformation of $x^T\beta$
$$ [Pr(x^T\beta)_L \leq Pr(x^T\beta) \leq Pr(x^T\beta)_U] = [F(x^T\beta)_L \leq F(x^T\beta) \leq F(x^T\beta)_U] $$
Concretely this means computing $\beta^Tx \pm z^* SE(\beta^Tx)$ and then applying the logit transform to the result to get the lower and upper bounds:
$$[\frac{e^{x^T\beta - z^* SE(x^T\beta)}}{1 + e^{x^T\beta - z^* SE(x^T\beta)}}, \frac{e^{x^T\beta + z^* SE(x^T\beta)}}{1 + e^{x^T\beta + z^* SE(x^T\beta)}},] $$
The estimated approximate variance of $x^T\beta$ can be calculated using the covariance matrix of the regression coefficients using
$$ Var(x^T\beta) = x^T \Sigma x$$
The advantage of this method is that the bounds cannot be outside the range $(0,1)$
There are several other approaches as well, using the delta method, bootstrapping etc.. which each have their own assumptions, advantages and limits.
Sources and info
My favorite book on this topic is "Applied Linear Statistical Models" by Kutner, Neter, Li, Chapter 14
Otherwise here are a few online sources:
Edit October 2021 - New links